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Akimi4 [234]
1 year ago
9

What is 4 principles of experimental design

Engineering
2 answers:
Mrrafil [7]1 year ago
7 0

Answer:

manipulation, control , random assignment, and random selection

Explanation:

tresset_1 [31]1 year ago
7 0

Answer:

Manipulation, control, random assignment, and random selection.

Explanation:

You might be interested in
Give two methods on how powder is produced in powder metallurgy.
Katen [24]

Answer:

Explanation:

  • <u>Atomization using gas stream</u>

Molten metal is forced through a small orifice and is shatter by a jet of compressed air,inert gas .

In Atomization, the particles shape is analysed  by the rate of solidification and varies from spherical to highly irregular shape.

  • Reduction

oxide of metals are transformed to pure metal powder when undefended to under melting point gases results in a product of spongy material.

It is used for Iron,copper,tungsten,Nickel etc.

7 0
4 years ago
Read 2 more answers
Create a C language program that can be used to construct any arbitrary Deterministic Finite Automaton corresponding to the FDA
otez555 [7]

Answer:

see the explanation

Explanation:

/* C Program to construct Deterministic Finite Automaton */

#include <stdio.h>

#include <DFA.h>

#include <stdlib.h>

#include <math.h>

#include <string.h>

#include <stdbool.h>

struct node{

struct node *initialStateID0;

struct node *presentStateID1;

};

printf("Please enter the total number of states:");

scanf("%d",&count);

//To create the Deterministic Finite Automata

DFA* create_dfa DFA(){

  q=(struct node *)malloc(sizeof(struct node)*count);

  dfa->initialStateID = -1;

  dfa->presentStateID = -1;

  dfa->totalNumOfStates = 0;

  return dfa;

}

//To make the next transition

void NextTransition(DFA* dfa, char c)

{

  int tID;

  for (tID = 0; tID < pPresentState->numOfTransitions; tID++){

       if (pPresentState->transitions[tID].condition(c))

      {

          dfa->presentStateID = pPresentState->transitions[tID].toStateID;

          return;

      }

  }

  dfa->presentStateID = pPresentState->defaultToStateID;

}

//To Add the state to DFA by using number of states

void State_add (DFA* pDFA, DFAState* newState)

{  

  newState->ID = pDFA->numOfStates;

  pDFA->states[pDFA->numOfStates] = newState;

  pDFA->numOfStates++;

}

void transition_Add (DFA* dfa, int fromStateID, int(*condition)(char), int toStateID)

{

  DFAState* state = dfa->states[fromStateID];

  state->transitions[state->numOfTransitions].toStateID = toStateID;

  state->numOfTransitions++;

}

void reset(DFA* dfa)

{

  dfa->presentStateID = dfa->initialStateID;

}

5 0
3 years ago
Solve the inequality. Then graph your solution.<br> -9v – 10 &lt; 7y +6
Cerrena [4.2K]
16

if you add 9+10 you get 18 - 7+6
5 0
3 years ago
A reservoir delivers water to a horizontal pipeline 39 long The first 15 m has a diameter of 50 mm, after which it suddenly beco
allsm [11]

Answer:

The difference of head in the level of reservoir is 0.23 m.

Explanation:

For pipe 1

d_1=50 mm,f_1=0.0048

For pipe 2

d_2=75 mm,f_2=0.0058

Q=2.8 l/s

Q=2.8\times 10^{-3]

We know that Q=AV

Q=A_1V_1=A_2V_2

A_1=1.95\times 10^{-3}m^2

A_2=4.38\times 10^{-3} m^2

So V_2=0.63 m/s,V_1=1.43 m/s

head loss (h)

h=\dfrac{f_1L_1V_1^2}{2gd_1}+\dfrac{f_2L_2V_2^2}{2gd_2}+0.5\dfrac{V_1^2}{2g}

Now putting the all values

h=\dfrac{0.0048\times 15\times 1.43^2}{2\times 9.81\times 0.05}+\dfrac{0.0058\times 24\times 0.63^2}{2\times 9.81\times 0.075}+0.5\dfrac{1.43^2}{2\times 9.81}

So h=0.23 m

So the difference of head in the level of reservoir is 0.23 m.

8 0
3 years ago
Find E[x] when x is sum of two fair dice?
Ksenya-84 [330]

Answer:

When two fair dice are rolled, 6×6=36 observations are obtained.

P(X=2)=P(1,1)=

36

1

​

P(X=3)=P(1,2)+P(2,1)=

36

2

​

=

18

1

​

P(X=4)=P(1,3)+P(2,2)+P(3,1)=

36

3

​

=

12

1

​

P(X=5)=P(1,4)+P(2,3)+P(3,2)+P(4,1)=

36

4

​

=

9

1

​

P(X=6)=P(1,5)+P(2,4)+P(3,3)+P(4,2)+P(5,1)=

36

5

​

P(X=7)=P(1,6)+P(2,5)+P(3,4)+P(4,3)+P(5,2)+P(6,1)=

36

6

​

=

6

1

​

P(X=8)=P(2,6)+P(3,5)+P(4,4)+P(5,3)+P(6,2)=

36

5

​

P(X=9)=P(3,6)+P(4,5)+P(5,4)+P(6,3)=

36

4

​

=

9

1

​

P(X=10)=P(4,6)+P(5,5)+P(6,4)=

36

3

​

=

12

1

​

P(X=11)=P(5,6)+P(6,5)=

36

2

​

=

18

1

​

P(X=12)=P(6,6)=

36

1

​

Therefore, the required probability distribution is as follows.

Then, E(X)=∑X

i

​

⋅P(X

i

​

)

=2×

36

1

​

+3×

18

1

​

+4×

12

1

​

+5×

9

1

​

+6×

36

5

​

+7×

6

1

​

+8×

36

5

​

+9×

9

1

​

+10×

12

1

​

+11×

18

1

​

+12×

36

1

​

=

18

1

​

+

6

1

​

+

3

1

​

+

9

5

​

+

6

5

​

+

6

7

​

+

9

10

​

+1+

6

5

​

+

18

11

​

+

3

1

​

=7

E(X

2

)=∑X

i

2

​

⋅P(X

i

​

)

=4×

36

1

​

+9×

18

1

​

+16×

12

1

​

+25×

9

1

​

+36×

36

5

​

+49×

6

1

​

+64×

36

5

​

+81×

9

1

​

+100×

12

1

​

+121×

18

1

​

+144×

36

1

​

=

9

1

​

+

2

1

​

+

3

4

​

+

9

25

​

+5+

6

49

​

+

9

80

​

+9+

3

25

​

+

18

121

​

+4

=

18

987

​

=

6

329

​

=54.833

Then, Var(X)=E(X

2

)−[E(X)]

2

=54.833−(7)

2

=54.833−49

=5.833

∴ Standard deviation =

Var(X)

​

=

5.833

​

=2.415

4 0
3 years ago
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